Math programming help for a gimble-based painting machine

Question

Hi -

I'm an artist involved with building various sorts of computer controlled machines. I've started prototyping a gimble-based XY painting machine and have realized that the maths needed are out of my reach. I'm a decent enough programmer but not strong in math- esp. 3D math.

To get a sense of what I'm needing to do, it might be helpful to look at the rig:

Early prototype:

http://roypardi.com/gimble/IMG_2803.JPG

http://roypardi.com/gimble/IMG_2805.JPG

http://roypardi.com/gimble/IMG_2806.JPG

http://roypardi.com/gimble/gimbleSmall.MOV (small video)

http://roypardi.com/gimble/gimbleLarge.mov (larger video)

The two inner rings represent the X/Y axes and are controlled by stepper motors. I want to be able to use both raster images and vector data (gcode). So I need to be able to address a point in 2D space on the paper/from my data and have the gimble figure out what orientation it needs to be at in order to get there (i.e. how much to step each motor).

I've been searching out 2D > 3D projection, Euler angles, etc. but I'm out of my depth. Any pointers, pushes in the right direction, or code snippets would be most welcome. I can make sense of most programming languages.

Thanks,

--Roy

Answer 1

I think it's a problem of simple http://en.wikipedia.org/wiki/Trigonometry

Let's say that the distance from the centre of your rings to the nearest point on the paper (which I'll call point 'O' for 'Origin') is distance X.

Take another point P directly north of O, whose distance from O is Y.

To paint this point, you need the angle alpha such that tan(alpha)=Y/X, i.e. you can calculate alpha using the formula "arctan(Y/X)" [arctan is sometimes also known as atan]. Arctan is a trignometric function, which I think you'll probably find defined in the API of a general purpose math library.

The above is the simplest case.

The only other case that I can think of is when the point P isn't due north. Instead of being due north, let's say that its distance is Y1 to the north, and Y2 to the east. The solution is two angles (one angle for each of two rings), one of which is "arctan(Y1/X)" and the other of which is "arctan(Y2/X)".

Answer 2

Very nice machine you have made, I hope this works for you I believe it is correct.

The way I see it, is to get one angle is simple, but the other is slightly harder to visualise as we have tilted the axis which it turns upon.

I'm going to avoid using tan, as when programming this could result in a division by 0, which could be frustrating. Also Z is going to be the height of the origin above the paper.

YAxis = arcsin( X / sqrt(X² + Z²))

XAxis = arcsin( Y / sqrt(Y² + X² + Z²))

or we could use

XAxis = arcsin(Y / sqrt(Y² + Z²))

YAxis = arcsin( X / sqrt(X² + Y² + Z²))

Also, I'd very much like to see a video of this plotting, if it works.

Edit: After thinking about it i believe only one solution will work it depends on which axis is affected by the other. Is the YAxis in the Middle or the Xaxis?

Answer 3

Perhaps I misunderstand, but I don't believe a gimbal will do what you want. A gimbal can point in any 3D direction, but it cannot move to arbitrary points in 3D space. If the plane of the paper intersects the volume swept by the pen held in the gimbal, the pen might be able to draw a circle, but nothing more. Even drawing a circle is not a sure thing, since in this case the paper would also intersect the volume swept by the gimbal rings; trying to orient the pen would make a ring hit the paper.

I think what you want is a plotter, not a gimbal.